Does the quadratic formula always work for a quadratic? If it does, why are the factors sometimes imaginary numbers?
[Math] Does the quadratic formula always work for a quadratic
quadratics
Related Solutions
Short answer: You have to remember, grouping is simply splitting the middle term so that it can be grouped:
Take the expression $ 3x^2+10x+8 $ The terms have no common factors, but if you just split the middle term right: $$3x^2+4x+6x+8$$ You can suddenly "group" the expression like such: $$x(3x+4)+2(3x+4)$$ Woah - Magic! Now we have $(x+2)(3x+4)$!
Long answer: Continuing off Gerry Myerson's answer:
Given the expression: $ ax^2+bx+c $ You could rewrite it as such: $(dx+e)(fx+g)$
Which proves the following is true (multiply it out): $$ ax^2+bx+c=dfx^2+(dg+ef)x+eg $$ What you want to do first is to find $dg$ and $ef$. Since you already know that $df = a$, and $eg =c$, all you have to do is factor $dfeg$, or $ac$, into $dg$ and $ef$ (remember, $dg+ef=b$). Once you have found $dg$ and $ef$, you can rewrite the expression as such: $$dfx^2+dgx+efx+eg $$ Now, you can "group" the equation into two factorable terms: $$dfx^2+dgx = dx(fx+g)$$$$ efx+eg=e(fx+g)$$ Yay! Now your expression is simply: $$dx(fx+eg)+e(fx+eg)$$$$=(dx+e)(fx+eg)$$
If you are only interested in solving an equation like $$x^2+3x+2=0$$ you are of course free to use any method you like (completing the square, use any formula, try to factorize...). But there is more to that than just solving equations.
If you solve the equation above you will get $x_1=-2,x_2=-1$. The fact that based on these solutions you can now write $x^2+3x+2=(x-x_1)(x-x_2)=(x+2)(x+1)$ holds for polynomials with higher degree. So if you solved the equation $$x^3+6x^2+11x+6=0$$ (which is a little bit trickier and would likely involve long division of polynomials), you'd get $x_1=-3,x_2=-2,x_1=-3$ and hence could write $$x^3+6x^2+11x+6=(x+3)(x+2)(x+1).$$ Just to outline some uses:
- Solving polynomial inequalities. Assume you are not interested in the quadratic equation $$x^2+3x+2=0$$ but in the quadratic inequation $$x^2+3x+2\geq 0.$$ There are no formulas to apply directly, so you have to use something different. Factorizing comes in handy as we can write: $$x^2+3x+2\geq 0 \Leftrightarrow (x+2)(x+1)\geq 0.$$ Now we know that the product of two (real) numbers is $\geq 0$, if and only if both numbers are $\geq 0$ or both numbers are $\leq 0$. Using this we can break that inequation down to two systems of linear inequations $$x+2\geq 0,~x+1\geq 0\quad \text{or}\quad x+2\leq0,~x+1\leq 0$$ which are easier to solve. Of course, this method can be refined and made more efficient, but it still needs you to factorize the given polynomial expression.
- Determining the type of a singularity. Let $$f:D\rightarrow\mathbb R,~f(x)=\frac{x^2+3x+2}{x^3+6x^2+11x+6}$$ where $D$ denotes the maximal domain of the function. As seen above we have $$x^3+6x^2+11x+6=0 \Leftrightarrow x\in\{-3,-2,-1\},$$ therefore we have $D=\mathbb R\setminus\{-3,-2,-1\}$. Now one can ask if there exists a contiuous extension of the function in $x=-3,-2,-1$. Using the factorizations from above we can write (for $x\neq -3,-2,-1$): $$f(x)=\frac{(x+2)(x+1)}{(x+3)(x+2)(x+1)}=\frac{1}{x+3}.$$ Thus we can say that there exists a continuous extension of $f$ in $x=-2$ and $x=-1$, but not in $x=-3$.
- Simplyfing expressions. More general than what I did in point 2, this can be used to simply expressions of the form $\frac{P(x)}{Q(x)}$ where $P,Q$ are polynomials. Uses of this are very wide (types of singularties of a function, calculating the limit of a sequence, haveing a "nicer" expression to differentiate...)
So it is not just another method to solve a equation in yet another, maybe time saving way, but it actually has some relevance for more advanced problems.
Best Answer
To answer your question, yes, the formula always works for quadratic equations, because from the equation $ax^2+bx+c=0$, one can derive the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ manually. Here is a video on Youtube showing such a derivation.
The factors become imaginary numbers if $b^2-4ac<0$, which means $b^2-4ac$ is negative. And taking the square root of a negative number produces an imaginary number.